Let f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} then \int e^x(f(x) + f'(x))dx (where c is the constant of integration)
Step-by-Step Solution
Key Concept: The integral is of the form integral e^x(f(x) + f'(x))dx = e^x f(x) + c. Simplify f(x) first.
f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{-(1 - 2\sin^2 x)}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = -\frac{\cos 2x}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}. Alternatively, simplify f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{2\sin^2 x - 1 + 2\sin x \cos^2 x + \cos^2 x}{\cos x(1 + \sin x)} = \frac{2\sin^2 x - 1 + 2\sin x(1 - \sin^2 x) + 1 - \sin^2 x}{\cos x(1 + \sin x)} = \frac{\sin^2 x + 2\sin x - 2\sin^3 x}{\cos x(1 + \sin x)} = \frac{\sin x(\sin x + 2 - 2\sin^2 x)}{\cos x(1 + \sin x)}. Actually, f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)(1 - \sin x)}{\cos^2 x} = \frac{2\sin^2 x - 1}{\cos x} + \frac{(2\sin x + 1)(1 - \sin x)}{\cos x} = \frac{2\sin^2 x - 1 + 2\sin x - 2\sin^2 x + 1 - \sin x}{\cos x} = \frac{\sin x}{\cos x} = \tan x. Thus, \int e^x(f(x) + f'(x))dx = e^x f(x) + c = e^x \tan x + c.
Correct Answer: 1